# proof of Cauchy-Schwarz inequality

If $a$ and $b$ are linearly dependent, we write $\bm{b}=\lambda \bm{a}$. So we get:

$${\u27e8\bm{a},\lambda \bm{a}\u27e9}^{2}={\lambda}^{2}{\u27e8\bm{a},\bm{a}\u27e9}^{2}={\lambda}^{2}{||\bm{a}||}^{4}={||\bm{a}||}^{2}{||\bm{b}||}^{2}.$$ |

So we have equality if $\bm{a}$ and $\bm{b}$ are linearly dependent. In the other case we look at the quadratic function

$${||x\cdot \bm{a}+\bm{b}||}^{2}={x}^{2}{||\bm{a}||}^{2}+2x\u27e8\bm{a},\bm{b}\u27e9+{||\bm{b}||}^{2}.$$ |

This function is positive for every real $x$, if $\bm{a}$ and $\bm{b}$ are linearly independent. Thus it has no real zeroes, which means that

$${\u27e8\bm{a},\bm{b}\u27e9}^{2}-{||\bm{a}||}^{2}{||\bm{b}||}^{2}$$ |

is always negative. So we have:

$$ |

which is the Cauchy-Schwarz inequality if $\bm{a}$ and $\bm{b}$ are linearly independent.

Title | proof of Cauchy-Schwarz inequality |
---|---|

Canonical name | ProofOfCauchySchwarzInequality |

Date of creation | 2013-03-22 12:34:42 |

Last modified on | 2013-03-22 12:34:42 |

Owner | mathwizard (128) |

Last modified by | mathwizard (128) |

Numerical id | 6 |

Author | mathwizard (128) |

Entry type | Proof |

Classification | msc 15A63 |